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# Whats the probability that in three rolls of a fair die, 6

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CEO
Joined: 15 Aug 2003
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Whats the probability that in three rolls of a fair die, 6 [#permalink]  30 Sep 2003, 12:44
Whats the probability that in three rolls of a fair die, 6 comes up at least twice.
Intern
Joined: 29 Aug 2003
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Location: Detroit, MI
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Re: Probability # 1 [#permalink]  01 Oct 2003, 05:52
Using Binomial Distribution,

The probability that an event will happen EXACTLY k times out of the n tries = nck * p^k * (1-p)^(n-k)

Probability of getting a 6 atleast 2 times = Probability of getting 6 exactly 2 times + Probability of getting 6 exactly 3 times

Probability of getting a 6 with a fair die = 1/6

Probability of getting 6 exactly 2 times = 3C2 (1/6)^2 + (1/6)^1 = 3/216
Probability of getting 6 exactly 3 times = 3C3 (1/6)^3 + (1/6)^0 = 1/216

So, Probability of getting a 6 atleast 2 times = 3/216 + 1/216 = 1/54

Am, I right?
Senior Manager
Joined: 21 Aug 2003
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Probability of 6 coming atleast two times = P(2) + P(3)
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36
Intern
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Vicky wrote:
Probability of 6 coming atleast two times = P(2) + P(3)
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36

Vicki,

In your notes, 5/216 is the probability of getting a 6 on the first 2 tries only, right? Should we not take in to account the cases where we can have 6's on the 2nd and 3rd tries and also 1st and 3rd tries?
Senior Manager
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yes amarsesh i made a mistake...
we should include getting 6 for 2nd, 3rd and 1st, 3rd time as well.
That will make it = 5*3/216 + 1/216 = 16/216 = 2/27
correct me if i am wrong.
thanks
Intern
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Location: Detroit, MI
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Vicky wrote:
yes amarsesh i made a mistake...
we should include getting 6 for 2nd, 3rd and 1st, 3rd time as well.
That will make it = 5*3/216 + 1/216 = 16/216 = 2/27
correct me if i am wrong.
thanks

Yes, you are right. I made a mistake in my calculations. I took 1/6 for the prob. that 6 is not obtained when I should have taken that as 5/6

It's obvious that I was not looking into my mistakes

Last edited by amarsesh on 02 Oct 2003, 12:06, edited 1 time in total.
CEO
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Re: Probability # 1 [#permalink]  02 Oct 2003, 12:00
i did the same mistake vicks did...but i think binomial is a great way to solve problems where we have a fixed # of trials and the prob of success and failure is known.. i will try to use it in future..

Three ways we can get two six's.

6 # 6
# 6 6
6 6 #

Two six's = 3* (1/6)^2 * 5/6 =15/216

Three six's = 1/6^3 = 1/216

Probability = 16/216 = 2/27

thanks
praetorian
Re: Probability # 1   [#permalink] 02 Oct 2003, 12:00
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